Multivariable calculus and vector analysis

Divergence theorem examples

Example 1

Compute ${\iint }_S\mathbf{F}\cdot d\mathbf{S}$ where

$$\begin{array}{lll}\hfill \mathbf{F}=\left(3x+z^{77},y^2-sin{x}^{2}z,xz+y{e}^{x^5}\right)& & \hfill \end{array}$$

and $S$ is surface of box

$$\begin{array}{lll}\hfill 0\le x\le 1,0\le y\le 3,0\le z\le 2.& & \hfill \end{array}$$

Use outward normal $\mathbf{n}$.

Solution: Given the ugly nature of the vector field, it would be hard to compute this integral directly. However, the divergence of $\mathbf{F}$ is nice:

$$\begin{array}{lll}\hfill \mathrm{div}\mathbf{F}=3+2y+x.& & \hfill \end{array}$$

We use the divergence theorem to convert the surface integral into a triple integral

$$\begin{array}{lll}\hfill {\iint }_S\mathbf{F}\cdot d\mathbf{S}={\iiint }_B\mathrm{div}\mathbf{F}dV& & \hfill \end{array}$$

where $B$ is the box

$$\begin{array}{lll}\hfill 0\le x\le 1,0\le y\le 3,0\le z\le 2.& & \hfill \end{array}$$

We compute the triple integral of $\mathrm{div}\mathbf{F}=3+2y+x$ over the box $B$:

$$\begin{array}{llll}\hfill {\iint }_S\mathbf{F}\cdot d\mathbf{S}& ={\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^3{\int \; <!--nolimits\; -->}_0^2\left(3+2y+x\right)dzdydx& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^3\left(6+4y+2x\right)dydx& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^1\left(18+18+6x\right)dx& \hfill & \\ \hfill & =36+3=39.& \hfill & \end{array}$$

Example 2

For $\mathbf{F}=\left(x{y}^2,y{z}^2,x^{2}z\right)$, use the divergence theorem to evaluate

$$\begin{array}{lll}\hfill {\iint }_S\mathbf{F}\cdot d\mathbf{S}& & \hfill \end{array}$$

where $S$ is the sphere of radius 3 centered at origin. Orient the surface with the outward pointing normal vector.

Solution: Since I am given a surface integral (over a closed surface) and told to use the divergence theorem, I must convert the surface integral into a triple integral over the region inside the surface.

Since $\mathrm{div}\mathbf{F}=y^2+z^2+x^2$, the surface integral is equal to the triple integral

$$\begin{array}{lll}\hfill {\iiint }_B\left(y^2+z^2+x^2\right)dV& & \hfill \end{array}$$

where $B$ is ball of radius 3.

To evaluate the triple integral, we can change variables to spherical coordinates. In spherical coordinates, the ball is

$$\begin{array}{lll}\hfill 0\le \rho \le 3,0\le \theta \le 2\pi ,0\le \varphi \le \pi .& & \hfill \end{array}$$

The integral is simply $x^2+y^2+z^2={\rho }^2$. For spherical coordinates, we know that the Jacobian determinant is $dV={\rho }^{2}sin\varphi d\varphi d\theta d\rho$. Therefore, the integral is

$$\begin{array}{llll}\hfill {\int \; <!--nolimits\; -->}_0^3{\int \; <!--nolimits\; -->}_0^{2\pi }{\int \; <!--nolimits\; -->}_0^{\pi }{\rho }^{4}sin\varphi d\varphi d\theta d\rho & ={\int \; <!--nolimits\; -->}_0^3{\int \; <!--nolimits\; -->}_0^{2\pi }{\left[-{\rho }^{4}cos\varphi \right]}_{\varphi =0}^{\varphi =\pi }d\theta d\rho & \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^3{\int \; <!--nolimits\; -->}_0^{2\pi }2{\rho }^{4}d\theta d\rho & \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^{3}4\pi {\rho }^{4}d\rho ={\left.\frac{4\pi {\rho }^5}{5}\right|}_0^3=\frac{972\pi }{5}.& \hfill & \end{array}$$

Contents by topic

1. Geometry in higher dimensions

2. Differential Calculus

3. Integral calculus

4. Vector calculus

5. The fundamental theorems of vector calculus

5a. Green's theorem

5b. The theorem for conservative vector fields

5c. Stokes' theorem

5d. The divergence theorem

• The idea of the divergence theorem
• Divergence theorem examples

6. Review material